There are 2011 positive numbers with both their sum and the sum of their reciprocals equal to 2012.  Let $x$ be one of these numbers.  Find the maximum value of $x + \frac{1}{x}.$
Solution: Let the other 2010 numbers be $y_1,$ $y_2,$ $\dots,$ $y_{2010}.$  Then $y_1 +y_2 + \dots + y_{2010} = 2012 - x$ and $\frac{1}{y_1} + \frac{1}{y_2} + \dots + \frac{1}{y_{2010}} = 2012 - \frac{1}{x}.$  By Cauchy-Schwarz,
\[\left( \sum_{i = 1}^{2010} y_i \right) \left( \sum_{i = 1}^{2010} \frac{1}{y_i} \right) = (2012 - x) \left( 2012 - \frac{1}{x} \right) \ge 2010^2.\]Then $2012^2 - 2012 \left( x + \frac{1}{x} \right) + 1 \ge 2010^2,$ which leads to
\[x + \frac{1}{x} \le \frac{8045}{2012}.\]The equation $x + \frac{1}{x} = \frac{8045}{2012}$ reduces to $x^2 - \frac{8045}{2012} x + 1 = 0,$ which has real roots.  We can then set $y_i = \frac{2012 - x}{2010}$ in order to achieve equality.  Thus, the maximum value is $\boxed{\frac{8045}{2012}}.$